How do we re-arrange a manifold cover without information on component-wise intersections in such a way that homology is computable?

Suppose a manifold is covered as the union of $|A|$-many $n$-spheres, i.e. $M=\bigcup_{\alpha\in A}S^n$. We are allowed to move the spheres around so long as they cover $M$; that is, we can make the intersection as small as possible (e.g. a point). I would like to compute the homology groups of $M$; however, we have no information regarding the intersections of of the spheres (other than the fact that they cannot be disjoint).

Is there a way to "arrange" the spheres in such a way that they still cover $M$ but have homology that is computable? Could I contract the intersection to a point and then compute the homology of a wedge sum $H_k\left(\bigvee_{\alpha\in A} S^n\right)=\bigoplus_{\alpha\in A} H_k(S^n)$?

Note, $A$ can be made to be countably infinite or uncountably infinite.

Thanks in advance! Any help would be much appreciated.

• I'm not sure I understand but when each $U_\alpha$ is homeomorphic to an open ball then the connected sum $U\alpha \#U_\beta$ is always a ball. So $\#U_\alpha$ would be a ball, right? Aug 22 '18 at 17:21
• At some point, you're going to have to attach a ball along something which is not a disk. Aug 22 '18 at 21:51
• @Multivariablecalculus: try your idea on a sphere, or even better, a circle. Aug 22 '18 at 21:59
• @Multivariablecalculus: Consider covering the circle with three overlapping coordinate charts (which are copies of open intervals.) Call them $U_1,U_2,U_3$ and assume $U_i\cap U_j$ is a small interval but the triple intersection is empty. Now you can say $U_1\cup U_2$ is homeomorphic to $U_1\#U_2$. However once you try to add $U_3$ in, it overlaps with $U_1\cup U_2$ in two different subintervals. By definition of connect sum, you can't attach to both simultaneously. Aug 23 '18 at 0:37
• @Multivariablecalculus: yes. You can build any smooth manifold up using a handle decomposition. The wikipedia article has some info, though I don't know how easy it is to understand if you've never seen it before. Aug 23 '18 at 0:44

• I have one final question. How do we do this explicitly for $\bigcup_{\alpha\in A}U_{\alpha}$ while retaining a cover of $M$? @CheerfulParsnip Aug 23 '18 at 17:49